Find all nonnegative real number $a$, such that $f(a)=0$ for any function $f$ satisfying: $xf(1+xf(y))=f(f(x)+f(y))$ with all $x,y$ are nonnegative real number.
I don't know why this problem only ask the number $a$ for $f(a)=0$ instead of the function $f(x)$. Is there any special thing here? Is it possible to show the bijection? I will show what I get from trying to solve this: $f(f(x)+f(0))=f(1+xf(0))=0$. How to find $t=f(0)$, show me your solution or any idea please, thank all.
I have started writing couse I thought I will solve it. But no, sorry. Perhaps you will get some idea from here. If you don't like it I will delete it.
We have: $$xf(1+xf(y))=f(f(x)+f(y))$$
Suppose there exists $a$ such that $f(a)=0$.
If $y=a$ we get: $bx = f(f(x))$ where $b=f(1)$. So if $\boxed{b\ne 0}$ then $f$ is injective (so $f$ has only one zero), so for $x=1$ we get $f(1+f(y)) = f(b+f(y))\implies b=1$. Now for $x=0$ we get $f(f(0)+f(y))=0$, so $f(0)+f(y)=a $ for all $y$, thus $f$ is constant. A contradiction. So $\boxed {b=0}$ and $f(f(x))=0$ for all $x$. So in special case (if we put $x=a$) we get $f(0)=0$.
If $x=a$ we get: $af(1+af(y)) = f(f(y))=0$